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Difference between revisions of "2003 AIME I Problems/Problem 4"
(prosify, box)
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Given that <math> \log_{10} \sin x + \log_{10} \cos x = -1 </math> and that <math> \log_{10} (\sin x + \cos x) = \frac{1}{2} (\log_{10} n - 1), </math> find <math> n. </math>
 
Given that <math> \log_{10} \sin x + \log_{10} \cos x = -1 </math> and that <math> \log_{10} (\sin x + \cos x) = \frac{1}{2} (\log_{10} n - 1), </math> find <math> n. </math>
 
== Solution ==
 
== Solution ==
−
<math> \log_{10} \sin x + \log_{10} \cos x = -1 </math>
+
The first equation, <math>\displaystyle \log_{10} \sin x + \log_{10} \cos x = -1 </math>, can be combined under the properties of [[logarithm]]s to <math> \displaystyle \log_{10}(\sin x \cos x) = -1 </math>. Therefore, <math> \sin x \cos x = \frac{1}{10} </math>.
  
−
<math> \log_{10}(\sin x \cos x) = -1 </math>
+
Now, manipulate the second equation, <math> \log_{10} (\sin x + \cos x) = \frac{1}{2} (\log_{10} n - 1) </math>.
  
−
<math> \sin x \cos x = \frac{1}{10} </math>
+
:<math> \log_{10} (\sin x + \cos x) = \frac{1}{2}(\log_{10} n - \log_{10} 10) </math>  
  
−
<math> \log_{10} (\sin x + \cos x) = \frac{1}{2} (\log_{10} n - 1) </math>
+
:<math> \log_{10} (\sin x + \cos x) = \frac{1}{2}(\log_{10} \frac{n}{10}) </math>  
  
−
<math> \log_{10} (\sin x + \cos x) = \frac{1}{2}(\log_{10} n - \log_{10} 10) </math>  
+
:<math> \log_{10} (\sin x + \cos x) = \left(\log_{10} \sqrt{\frac{n}{10}}\right) </math>  
  
−
<math> \log_{10} (\sin x + \cos x) = \frac{1}{2}(\log_{10} \frac{n}{10}) </math>  
+
:<math> \sin x + \cos x = \sqrt{\frac{n}{10}} </math>
  
−
<math> \log_{10} (\sin x + \cos x) = (\log_{10} \sqrt{\frac{n}{10}}) </math>  
+
:<math> (\sin x + \cos x)^{2} = \left(\sqrt{\frac{n}{10}}\right)^2 </math>
  
−
<math> \sin x + \cos x = \sqrt{\frac{n}{10}} </math>
+
:<math> \sin^2 x + \cos^2 x +2 \sin x \cos x= \frac{n}{10} </math>
  
−
<math> (\sin x + \cos x)^{2} = (\sqrt{\frac{n}{10}})^2 </math>
+
<math>\displaystyle \sin ^2 x + \cos ^2 x = 1</math>, and we can substitute the value for <math>\displaystyle \sin x \cos x</math> from above.
  
−
<math> \sin^2 x + \cos^2 x +2 \sin x \cos x= \frac{n}{10} </math>
+
:<math> 1 + 2(\frac{1}{10}) = \frac{n}{10} </math>
  
−
<math> 1 + 2(\frac{1}{10}) = \frac{n}{10} </math>
+
:<math> \frac{12}{10} = \frac{n}{10} </math>
  
−
<math> \frac{12}{10} = \frac{n}{10} </math>
+
Thus, the solution is <math> n = 012 </math>.
−
 
 
−
<math> n = 012 </math>
 
  
 
== See also ==
 
== See also ==
−
* [[2003 AIME I Problems/Problem 3|Previous Problem]]
 
−
* [[2003 AIME I Problems/Problem 5|Next Problem]]
 
−
* [[2003 AIME I Problems]]
 
 
* [[Logarithm]]
 
* [[Logarithm]]
 
* [[Trigonometry]]
 
* [[Trigonometry]]
 +
{{AIME box|year=2003|n=I|num-b=3|num-a=5}}
  
 
[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Trigonometry Problems]]
 
[[Category:Intermediate Trigonometry Problems]]

Revision as of 16:48, 8 March 2007

Problem

Given that $\log_{10} \sin x + \log_{10} \cos x = -1$ and that $\log_{10} (\sin x + \cos x) = \frac{1}{2} (\log_{10} n - 1),$ find $n.$

Solution

The first equation, $\displaystyle \log_{10} \sin x + \log_{10} \cos x = -1$, can be combined under the properties of logarithms to $\displaystyle \log_{10}(\sin x \cos x) = -1$. Therefore, $\sin x \cos x = \frac{1}{10}$.

Now, manipulate the second equation, $\log_{10} (\sin x + \cos x) = \frac{1}{2} (\log_{10} n - 1)$.

$\log_{10} (\sin x + \cos x) = \frac{1}{2}(\log_{10} n - \log_{10} 10)$
$\log_{10} (\sin x + \cos x) = \frac{1}{2}(\log_{10} \frac{n}{10})$
$\log_{10} (\sin x + \cos x) = \left(\log_{10} \sqrt{\frac{n}{10}}\right)$
$\sin x + \cos x = \sqrt{\frac{n}{10}}$
$(\sin x + \cos x)^{2} = \left(\sqrt{\frac{n}{10}}\right)^2$
$\sin^2 x + \cos^2 x +2 \sin x \cos x= \frac{n}{10}$

$\displaystyle \sin ^2 x + \cos ^2 x = 1$, and we can substitute the value for $\displaystyle \sin x \cos x$ from above.

$1 + 2(\frac{1}{10}) = \frac{n}{10}$
$\frac{12}{10} = \frac{n}{10}$

Thus, the solution is $n = 012$.

See also

2003 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
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All AIME Problems and Solutions
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